white-box testing techniques iiwhite-box testing topics •logic coverage (lecture i) •dataflow...

White-Box Testing Techniques II

Prepared by

Stephen M. Thebaut, Ph.D.

University of Florida

Software Testing and Verification

Lecture 8

White-Box Testing Topics

• Logic coverage (lecture I)

• Dataflow coverage (lecture II)

• Path conditions and symbolic execution (lecture III)

• Other white-box testing strategies (e.g., “fault-based testing”) (lecture IV)

Dataflow Coverage

• Based on the idea that program paths along which variables are defined and then used should be covered.

• Families of path selection criteria have been defined based on somewhat different models/definitions. The different criteria reflect different degrees of coverage.

• The model/definitions we consider are representative.

• CASE tool support is very desirable.

Dataflow Coverage

Variable Definitions and Uses

• A program variable is DEFINED when it appears:

– on the left hand side of an assignment statement (e.g., Y := 17)

– in an input statement (e.g., input(Y))

– as an OUT parameter in a subroutine call (e.g., DOIT(X:IN,Y:OUT))

(cont’d)

• A program variable is USED when it appears:

– on the right hand side of an assignment statement (e.g., Y := X+17)

– as an IN parameter in a subroutine or function call (e.g., Y := SQRT(X))

– in the predicate of a branch statement (e.g., if X>0 then...)

(cont’d)

• Use of a variable in the predicate of a branch statement is called a predicate-use (“p-use”). Any other use is called a computation-use (“c-use”).

• For example, in the program statement:

If (X>0) then

print(Y)

end_if_then

there is a p-use of X and a c-use of Y.

(cont’d)

• Use of a variable in the predicate of a branch statement is called a predicate-use (“p-use”). Any other use is called a computation-use (“c-use”).

• For example, in the program statement:

If (X>0) then

print(Y)

end_if_then

there is a p-use of X and a c-use of Y.

(cont’d)

• A variable can also be used and then re-defined in a single statement when it appears:

– on both sides of an assignment statement (e.g., Y := Y+X)

– as an IN/OUT parameter in a subroutine call (e.g., INCREMENT(Y:IN/OUT))

(cont’d)

Other Dataflow Terms and (Informal)

Definitions

• A path is definition clear (“def-clear”)with respect to a variable v if there is no re-definition of v along the path.

• A complete path is a path whose initial node is a/the start node and whose final node is a/the exit node.

Definitions

• A path is definition clear (“def-clear”)with respect to a variable v if there is no re-definition of v along the path.

• A complete path is a path whose initial node is a/the start node and whose final node is a/the exit node.

(cont’d)

Definitions (cont’d)

• A definition-use pair (“du-pair”) with respect to a variable v is a double (d,u) such that d is a node in the program’s flow graph at which v is defined, u is a node or edge at which v is used, and there is a def-clear path with respect to v from d to u.

• (Note that the definition of a du-pair does not require the existence of a feasible def-clear path from d to u.)

Example 1

1. input(A,B)

if (B>1) then

2. A := A+7

end_if

3. if (A>10) then

4. B := A+B

end_if

5. output(A,B)

input(A,B)

A := A+7

B := A+B

output(A,B)

Identifying DU-Pairs – Variable A

du-pair path(s)

(1,2) <1,2>

(1,4) <1,3,4>

(1,5) <1,3,4,5>

<1,3,5>

(1,<3,4>) <1,3,4>

(1,<3,5>) <1,3,5>

(2,4) <2,3,4>

(2,5) <2,3,4,5>

<2,3,5>

(2,<3,4>) <2,3,4>

(2,<3,5>) <2,3,5>

input(A,B)

A := A+7

B := A+B

output(A,B)

du-pair path(s)

(1,2) <1,2>

(1,4) <1,3,4>

(1,5) <1,3,4,5>

<1,3,5>

(1,<3,4>) <1,3,4>

(1,<3,5>) <1,3,5>

(2,4) <2,3,4>

(2,5) <2,3,4,5>

<2,3,5>

(2,<3,4>) <2,3,4>

(2,<3,5>) <2,3,5>

input(A,B)

A := A+7

B := A+B

output(A,B)

du-pair path(s)

(1,2) <1,2>

(1,4) <1,3,4>

(1,5) <1,3,4,5>

<1,3,5>

(1,<3,4>) <1,3,4>

(1,<3,5>) <1,3,5>

(2,4) <2,3,4>

(2,5) <2,3,4,5>

<2,3,5>

(2,<3,4>) <2,3,4>

(2,<3,5>) <2,3,5>

input(A,B)

A := A+7

B := A+B

output(A,B)

du-pair path(s)

(1,2) <1,2>

(1,4) <1,3,4>

(1,5) <1,3,4,5>

<1,3,5>

(1,<3,4>) <1,3,4>

(1,<3,5>) <1,3,5>

(2,4) <2,3,4>

(2,5) <2,3,4,5>

<2,3,5>

(2,<3,4>) <2,3,4>

(2,<3,5>) <2,3,5>

input(A,B)

A := A+7

B := A+B

output(A,B)

du-pair path(s)

(1,2) <1,2>

(1,4) <1,3,4>

(1,5) <1,3,4,5>

<1,3,5>

(1,<3,4>) <1,3,4>

(1,<3,5>) <1,3,5>

(2,4) <2,3,4>

(2,5) <2,3,4,5>

<2,3,5>

(2,<3,4>) <2,3,4>

(2,<3,5>) <2,3,5>

input(A,B)

A := A+7

B := A+B

output(A,B)

du-pair path(s)

(1,2) <1,2>

(1,4) <1,3,4>

(1,5) <1,3,4,5>

<1,3,5>

(1,<3,4>) <1,3,4>

(1,<3,5>) <1,3,5>

(2,4) <2,3,4>

(2,5) <2,3,4,5>

<2,3,5>

(2,<3,4>) <2,3,4>

(2,<3,5>) <2,3,5>

input(A,B)

A := A+7

B := A+B

output(A,B)

du-pair path(s)

(1,2) <1,2>

(1,4) <1,3,4>

(1,5) <1,3,4,5>

<1,3,5>

(1,<3,4>) <1,3,4>

(1,<3,5>) <1,3,5>

(2,4) <2,3,4>

(2,5) <2,3,4,5>

<2,3,5>

(2,<3,4>) <2,3,4>

(2,<3,5>) <2,3,5>

input(A,B)

A := A+7

B := A+B

output(A,B)

du-pair path(s)

(1,2) <1,2>

(1,4) <1,3,4>

(1,5) <1,3,4,5>

<1,3,5>

(1,<3,4>) <1,3,4>

(1,<3,5>) <1,3,5>

(2,4) <2,3,4>

(2,5) <2,3,4,5>

<2,3,5>

(2,<3,4>) <2,3,4>

(2,<3,5>) <2,3,5>

input(A,B)

A := A+7

B := A+B

output(A,B)

du-pair path(s)

(1,2) <1,2>

(1,4) <1,3,4>

(1,5) <1,3,4,5>

<1,3,5>

(1,<3,4>) <1,3,4>

(1,<3,5>) <1,3,5>

(2,4) <2,3,4>

(2,5) <2,3,4,5>

<2,3,5>

(2,<3,4>) <2,3,4>

(2,<3,5>) <2,3,5>

input(A,B)

A := A+7

B := A+B

output(A,B)

du-pair path(s)

(1,2) <1,2>

(1,4) <1,3,4>

(1,5) <1,3,4,5>

<1,3,5>

(1,<3,4>) <1,3,4>

(1,<3,5>) <1,3,5>

(2,4) <2,3,4>

(2,5) <2,3,4,5>

<2,3,5>

(2,<3,4>) <2,3,4>

(2,<3,5>) <2,3,5>

input(A,B)

A := A+7

B := A+B

output(A,B)

du-pair path(s)

(1,2) <1,2>

(1,4) <1,3,4>

(1,5) <1,3,4,5>

<1,3,5>

(1,<3,4>) <1,3,4>

(1,<3,5>) <1,3,5>

(2,4) <2,3,4>

(2,5) <2,3,4,5>

<2,3,5>

(2,<3,4>) <2,3,4>

(2,<3,5>) <2,3,5>

input(A,B)

A := A+7

B := A+B

output(A,B)

du-pair path(s)

(1,2) <1,2>

(1,4) <1,3,4>

(1,5) <1,3,4,5>

<1,3,5>

(1,<3,4>) <1,3,4>

(1,<3,5>) <1,3,5>

(2,4) <2,3,4>

(2,5) <2,3,4,5>

<2,3,5>

(2,<3,4>) <2,3,4>

(2,<3,5>) <2,3,5>

input(A,B)

A := A+7

B := A+B

output(A,B)

Identifying DU-Pairs – Variable B

du-pair path(s)

(1,4) <1,2,3,4>

<1,3,4>

(1,5) <1,2,3,5>

<1,3,5>

(1,<1,2>) <1,2>

(1,<1,3>) <1,3>

(4,5) <4,5>

input(A,B)

A := A+7

B := A+B

output(A,B)

(Verification of du-pairs/paths(s) left as an exercise...)

Dataflow Test Coverage Criteria

• All-Defs: for every program variable v, at least one def-clear path from every definition of v to at least one c-use or one p-use of v must be covered.

(cont’d)

• Consider a test case executing path:

1. <1,2,3,4,5>

• Identify all def-clear paths (associated with du-pairs) covered (i.e., subsumed) by this test case for each variable.

• Are all definitions for each variable associated with at least one of the subsumed def-clear paths?

(cont’d)

1. <1,2,3,4,5>

• Identify all def-clear paths (associated with du-pairs) covered (i.e., subsumed)by this test case for each variable.

• Are all definitions for each variable associated with at least one of the subsumed def-clear paths?

(cont’d)

1. <1,2,3,4,5>

• Identify all def-clear paths (associated with du-pairs) covered (i.e., subsumed)by this test case for each variable.

• Are all definitions for each variableassociated with at least one of the subsumed def-clear paths?

Def-Clear Paths Subsumed by <1,2,3,4,5>

for Variable A

du-pair path(s)

(1,2) <1,2>

(1,4) <1,3,4>

(1,5) <1,3,4,5>

<1,3,5>

(1,<3,4>) <1,3,4>

(1,<3,5>) <1,3,5>

(2,4) <2,3,4>

(2,5) <2,3,4,5>

<2,3,5>

(2,<3,4>) <2,3,4>

(2,<3,5>) <2,3,5>

input(A,B)

A := A+7

B := A+B

output(A,B)

Def-Clear Paths Subsumed by <1,2,3,4,5>

for Variable B

du-pair path(s)

(1,4) <1,2,3,4>

<1,3,4>

(1,5) <1,2,3,5>

<1,3,5>

(1,<1,2>) <1,2>

(1,<1,3>) <1,3>

(4,5) <4,5>

input(A,B)

A := A+7

B := A+B

output(A,B)

(cont’d)

• Since <1,2,3,4,5> covers at least one def-

clear path from each of the two definitions of A/B to at least one c-use or p-use of A/B, All-Defs coverage is achieved.

(cont’d)

• All-Uses: for every program variable v, at least one def-clear path from every definition of v to every c-useand every p-use of v must be covered.

• Consider additional test cases executing paths:

2. <1,3,4,5>

3. <1,2,3,5>

• Do all three test cases provide All-Uses coverage?

(cont’d)

2. <1,3,4,5>

3. <1,2,3,5>

(cont’d)

2. <1,3,4,5>

3. <1,2,3,5>

Def-Clear Paths Subsumed by <1,3,4,5>

for Variable A

du-pair path(s)

(1,2) <1,2>

(1,4) <1,3,4>

(1,5) <1,3,4,5>

<1,3,5>

(1,<3,4>) <1,3,4>

(1,<3,5>) <1,3,5>

(2,4) <2,3,4>

(2,5) <2,3,4,5>

<2,3,5>

(2,<3,4>) <2,3,4>

(2,<3,5>) <2,3,5>

input(A,B)

A := A+7

B := A+B

output(A,B)

for Variable B

du-pair path(s)

(1,4) <1,2,3,4>

<1,3,4>

(1,5) <1,2,3,5>

<1,3,5>

(1,<1,2>) <1,2>

(1,<1,3>) <1,3>

(4,5) <4,5>

input(A,B)

A := A+7

B := A+B

output(A,B)

for Variable A

du-pair path(s)

(1,2) <1,2>

(1,4) <1,3,4>

(1,5) <1,3,4,5>

<1,3,5>

(1,<3,4>) <1,3,4>

(1,<3,5>) <1,3,5>

(2,4) <2,3,4>

(2,5) <2,3,4,5>

<2,3,5>

(2,<3,4>) <2,3,4>

(2,<3,5>) <2,3,5>

input(A,B)

A := A+7

B := A+B

output(A,B)

for Variable B

input(A,B)

A := A+7

B := A+B

output(A,B)

du-pair path(s)

(1,4) <1,2,3,4>

<1,3,4>

(1,5) <1,2,3,5>

<1,3,5>

(1,<1,2>) <1,2>

(1,<1,3>) <1,3>

(4,5) <4,5>

(cont’d)

• Since none of the three test cases covers the du-pair (1,<3,5>) for variable A, All-Uses Coverage is not achieved.

• Exercise: Did the three test cases considered provide Branch coverage? Basis Path coverage? Path coverage?

(cont’d)

• Since none of the three test cases covers the du-pair (1,<3,5>) for variable A, All-Uses Coverage is not achieved.

• Exercise: Did the three test cases considered provide Branch coverage? Basis Paths coverage? Path coverage?

What would be even stronger than All-Uses?

• We have considered the All-Defs and the All-Uses criteria so far. How could the definition of All-Uses be modified to make it even stronger? Recall:

All-Uses: for every program variable v, at least one def-clear path from every definition of v to every c-use and every p-use of v must be covered.

What would be even stronger than All-Uses?

• We have considered the All-Defs and the All-Uses criteria so far. How could the definition of All-Uses be modified to make it even stronger? Recall:

All-Uses: for every program variable v, at least one def-clear path from every definition of v to every c-use and every p-use of v must be covered.

Example 2

1. input(X,Y)

2. while (Y>0) do

3. if (X>0) then

4. Y := Y-X

5. input(X)

end_if_then_else

6. end_while

7. output(X,Y)

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

Example 2

1. input(X,Y)

2. while (Y>0) do

3. if (X>0) then

4. Y := Y-X

5. input(X)

end_if_then_else

6. end_while

7. output(X,Y)

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

Note edges <6,7> and <6,3>.

Logic Coverage Control Flow Graph

input(a)

while a do

if b then s1

else s2

end_if_then_else

end_while

output(a)

Dataflow Coverage Control Flow Graph

input(a)

while a do

if b then s1

else s2

end_if_then_else

end_while

output(a)

Identifying DU-Pairs – Variable X

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(1,4) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,7) <1,2,7>

<1,2,3,4,6,7>

<1,2,3,4,6,(3,4,6)*,7>

(1,<3,4>) <1,2,3,4>

<1,2,3,4,(6,3,4)*>

(1,<3,5>) <1,2,3,5>

(5,4) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

(cont’d)

du-pair path(s)

(5,7) <5,6,7>

<5,6,3,4,6,7>

<5,6,3,4,6,(3,4,6)*,7>

(5,<3,4>) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

(5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5>

<5,6,3,4,6,(3,4,6)*,3,5>

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

du-pair path(s)

(5,7) <5,6,7> †

<5,6,3,4,6,7>

<5,6,3,4,6,(3,4,6)*,7>

(5,<3,4>) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

(5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5>

<5,6,3,4,6,(3,4,6)*,3,5>

† infeasible

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

du-pair path(s)

(5,7) <5,6,7> †

<5,6,3,4,6,7>

<5,6,3,4,6,(3,4,6)*,7>

(5,<3,4>) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

(5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5>

<5,6,3,4,6,(3,4,6)*,3,5>

† infeasible

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

du-pair path(s)

(5,7) <5,6,7> †

<5,6,3,4,6,7>

<5,6,3,4,6,(3,4,6)*,7>

(5,<3,4>) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

(5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5>

<5,6,3,4,6,(3,4,6)*,3,5>

† infeasible

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

du-pair path(s)

(5,7) <5,6,7> †

<5,6,3,4,6,7>

<5,6,3,4,6,(3,4,6)*,7>

(5,<3,4>) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

(5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5>

<5,6,3,4,6,(3,4,6)*,3,5>

† infeasible

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

du-pair path(s)

(5,7) <5,6,7> †

<5,6,3,4,6,7>

<5,6,3,4,6,(3,4,6)*,7>

(5,<3,4>) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

(5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5>

<5,6,3,4,6,(3,4,6)*,3,5>

† infeasible

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

du-pair path(s)

(5,7) <5,6,7> †

<5,6,3,4,6,7>

<5,6,3,4,6,(3,4,6)*,7>

(5,<3,4>) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

(5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5>

<5,6,3,4,6,(3,4,6)*,3,5>

† infeasible

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

du-pair path(s)

(5,7) <5,6,7> †

<5,6,3,4,6,7>

<5,6,3,4,6,(3,4,6)*,7>

(5,<3,4>) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

(5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5>

<5,6,3,4,6,(3,4,6)*,3,5>

† infeasible

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

du-pair path(s)

(5,7) <5,6,7> †

<5,6,3,4,6,7>

<5,6,3,4,6,(3,4,6)*,7>

(5,<3,4>) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

(5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5>

<5,6,3,4,6,(3,4,6)*,3,5>

† infeasible

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

du-pair path(s)

(5,7) <5,6,7> †

<5,6,3,4,6,7>

<5,6,3,4,6,(3,4,6)*7>

(5,<3,4>) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

(5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5>

<5,6,3,4,6,(3,4,6)*,3,5>

† infeasible

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

du-pair path(s)

(5,7) <5,6,7> †

<5,6,3,4,6,7>

<5,6,3,4,6,(3,4,6)*7>

(5,<3,4>) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

(5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5>

<5,6,3,4,6,(3,4,6)*,3,5>

† infeasible

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

du-pair path(s)

(5,7) <5,6,7> †

<5,6,3,4,6,7>

<5,6,3,4,6,(3,4,6)*7>

(5,<3,4>) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

(5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5> †

<5,6,3,4,6,(3,4,6)*,3,5>

† infeasible

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

du-pair path(s)

(5,7) <5,6,7> †

<5,6,3,4,6,7>

<5,6,3,4,6,(3,4,6)*7>

(5,<3,4>) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

(5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5> †

<5,6,3,4,6,(3,4,6)*,3,5> †

† infeasible

input(X,Y))

Y := Y-Xinput(X)

output(X,Y)

More Dataflow Terms and Definitions

• A path (either partial or complete) is simple if all edges within the path are distinct (i.e., different).

• A path is loop-free if all nodes within the path are distinct (i.e., different).

• A path (either partial or complete) is simple if all edges within the path are distinct (i.e., different).

• A path is loop-free if all nodes within the path are distinct (i.e., different).

Simple and Loop-Free Paths

path Simple? Loop-free?

<1,3,4,2> ? ?

<1,2,3,2>

<1,2,3,1,2>

<1,2,3,2,4>

<1,3,4,2> yes yes

<1,2,3,2>

<1,2,3,1,2>

<1,2,3,2,4>

<1,3,4,2> yes yes

<1,2,3,2> ? ?

<1,2,3,1,2>

<1,2,3,2,4>

<1,3,4,2> yes yes

<1,2,3,2> yes no

<1,2,3,1,2>

<1,2,3,2,4>

<1,3,4,2> yes yes

<1,2,3,2> yes no

<1,2,3,1,2> ? ?

<1,2,3,2,4>

<1,3,4,2> yes yes

<1,2,3,2> yes no

<1,2,3,1,2> no no

<1,2,3,2,4>

<1,3,4,2> yes yes

<1,2,3,2> yes no

<1,2,3,1,2> no no

<1,2,3,2,4> ? ?

<1,3,4,2> yes yes

<1,2,3,2> yes no

<1,2,3,1,2> no no

<1,2,3,2,4> yes no

Simple and Loop-Free Paths (cont’d)

Which is stronger, simple or loop-free?

Loop-free => Simple

Simple => Loop-free

A path <n1,n2,...,nj,nk> is a du-path

with respect to a variable v if v is defined at node n1 and either:

1. there is a c-use of v at node nk and<n1,n2,...,nj,nk> is a def-clear simple path, or

2. there is a p-use of v at edge <nj,nk> and <n1,n2,...nj> is a def-clear loop-

free path.

A path <n1,n2,...,nj,nk> is a du-path

with respect to a variable v if v is defined at node n1 and either:

1. there is a c-use of v at node nk and<n1,n2,...,nj,nk> is a def-clear simple path, or

2. there is a p-use of v at edge <nj,nk> and <n1,n2,...nj> is a def-clear loop-

free path.

Identifying du-paths

du-pair path(s) du-path?

X: (5,7) <5,6,7> † ?

<5,6,3,4,6,7>

<5,6,3,4,6,(3,4,6)*,7>

X: (5,<3,4>) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

X: (5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5> †

<5,6,3,4,6,(3,4,6)*,3,5> †

† infeasible

X: (5,7) <5,6,7> † yes

<5,6,3,4,6,7>

<5,6,3,4,6,(3,4,6)*,7>

X: (5,<3,4>) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

X: (5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5> †

<5,6,3,4,6,(3,4,6)*,3,5> †

† infeasible

X: (5,7) <5,6,7> † yes

<5,6,3,4,6,7> ?

<5,6,3,4,6,(3,4,6)*,7>

X: (5,<3,4>) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

X: (5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5> †

<5,6,3,4,6,(3,4,6)*,3,5> †

† infeasible

X: (5,7) <5,6,7> † yes

<5,6,3,4,6,7> yes

<5,6,3,4,6,(3,4,6)*,7>

X: (5,<3,4>) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

X: (5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5> †

<5,6,3,4,6,(3,4,6)*,3,5> †

† infeasible

X: (5,7) <5,6,7> † yes

<5,6,3,4,6,7> yes

<5,6,3,4,6,(3,4,6)*,7> ?

X: (5,<3,4>) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

X: (5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5> †

<5,6,3,4,6,(3,4,6)*,3,5> †

† infeasible

X: (5,7) <5,6,7> † yes

<5,6,3,4,6,7> yes

<5,6,3,4,6,(3,4,6)*,7> no

X: (5,<3,4>) <5,6,3,4>

<5,6,3,4,(6,3,4)*>

X: (5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5> †

<5,6,3,4,6,(3,4,6)*,3,5> †

† infeasible

X: (5,7) <5,6,7> † yes

<5,6,3,4,6,7> yes

<5,6,3,4,6,(3,4,6)*,7> no

X: (5,<3,4>) <5,6,3,4> ?

<5,6,3,4,(6,3,4)*>

X: (5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5> †

<5,6,3,4,6,(3,4,6)*,3,5> †

† infeasible

X: (5,7) <5,6,7> † yes

<5,6,3,4,6,7> yes

<5,6,3,4,6,(3,4,6)*,7> no

X: (5,<3,4>) <5,6,3,4> yes

<5,6,3,4,(6,3,4)*>

X: (5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5> †

<5,6,3,4,6,(3,4,6)*,3,5> †

† infeasible

X: (5,7) <5,6,7> † yes

<5,6,3,4,6,7> yes

<5,6,3,4,6,(3,4,6)*,7> no

X: (5,<3,4>) <5,6,3,4> yes

<5,6,3,4,(6,3,4)*> ?

X: (5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5> †

<5,6,3,4,6,(3,4,6)*,3,5> †

† infeasible

X: (5,7) <5,6,7> † yes

<5,6,3,4,6,7> yes

<5,6,3,4,6,(3,4,6)*,7> no

X: (5,<3,4>) <5,6,3,4> yes

<5,6,3,4,(6,3,4)*> no

X: (5,<3,5>) <5,6,3,5>

<5,6,3,4,6,3,5> †

<5,6,3,4,6,(3,4,6)*,3,5> †

† infeasible

X: (5,7) <5,6,7> † yes

<5,6,3,4,6,7> yes

<5,6,3,4,6,(3,4,6)*,7> no

X: (5,<3,4>) <5,6,3,4> yes

<5,6,3,4,(6,3,4)*> no

X: (5,<3,5>) <5,6,3,5> ?

<5,6,3,4,6,3,5> †

<5,6,3,4,6,(3,4,6)*,3,5> †

† infeasible

X: (5,7) <5,6,7> † yes

<5,6,3,4,6,7> yes

<5,6,3,4,6,(3,4,6)*,7> no

X: (5,<3,4>) <5,6,3,4> yes

<5,6,3,4,(6,3,4)*> no

X: (5,<3,5>) <5,6,3,5> yes

<5,6,3,4,6,3,5> †

<5,6,3,4,6,(3,4,6)*,3,5> †

† infeasible

X: (5,7) <5,6,7> † yes

<5,6,3,4,6,7> yes

<5,6,3,4,6,(3,4,6)*,7> no

X: (5,<3,4>) <5,6,3,4> yes

<5,6,3,4,(6,3,4)*> no

X: (5,<3,5>) <5,6,3,5> yes

<5,6,3,4,6,3,5> † ?

<5,6,3,4,6,(3,4,6)*,3,5> †

† infeasible

X: (5,7) <5,6,7> † yes

<5,6,3,4,6,7> yes

<5,6,3,4,6,(3,4,6)*,7> no

X: (5,<3,4>) <5,6,3,4> yes

<5,6,3,4,(6,3,4)*> no

X: (5,<3,5>) <5,6,3,5> yes

<5,6,3,4,6,3,5> † no

<5,6,3,4,6,(3,4,6)*,3,5> †

† infeasible

X: (5,7) <5,6,7> † yes

<5,6,3,4,6,7> yes

<5,6,3,4,6,(3,4,6)*,7> no

X: (5,<3,4>) <5,6,3,4> yes

<5,6,3,4,(6,3,4)*> no

X: (5,<3,5>) <5,6,3,5> yes

<5,6,3,4,6,3,5> † no

<5,6,3,4,6,(3,4,6)*,3,5> † ?

† infeasible

X: (5,7) <5,6,7> † yes

<5,6,3,4,6,7> yes

<5,6,3,4,6,(3,4,6)*,7> no

X: (5,<3,4>) <5,6,3,4> yes

<5,6,3,4,(6,3,4)*> no

X: (5,<3,5>) <5,6,3,5> yes

<5,6,3,4,6,3,5> † no

<5,6,3,4,6,(3,4,6)*,3,5> † no

† infeasible

Another Dataflow Test Coverage

Criterion

• All-DU-Paths: for every program variable v, every du-path from every definition of v to every c-use and every p-use of v must be covered.

Exercise

• Identify all c-uses and p-uses for variable Y in Example 2.

• For each c-use or p-use, identify (using the “ * ” notation) all def-clear paths.

• Identify whether or not each def-clear path is feasible, and whether or not it is a du-path.

Exercise

Summary of White-Box Coverage

RelationshipsPath

All du-paths

All-Uses

All-Defs

Compound Condition

Branch / Condition Basis Paths Loop

Condition

Statement

Branch

- nominal cases*

White-Box Testing Techniques II

Prepared by

Stephen M. Thebaut, Ph.D.

University of Florida

Software Testing and Verification

Lecture 8

white-box testing techniques iiwhite-box testing topics •logic coverage (lecture i) •dataflow...

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